fluid machinery

1. ME 411 – Fluid Machinery
Air Compressor
Eva O. Buenafe
ME - 3209

2. Air Compressors
An air compressor is a mechanical device that converts
power from a source, such as an electric motor or a gasoline
engine, into potential energy stored in compressed air. It
essentially compresses and pressurizes air to a higher pressure
than the surrounding atmospheric pressure.
MAIN COMPONENTS
1. a motor or engine,
2. a compression chamber
3. a tank

3. Air Compressors
Air Compressor
1. Takes in atmospheric air,
2. Compresses it, and
3. Delivers it to a storage vessel ( i.e. Reservoir ).
• Air compressors are used in a wide range of applications across
various industries. They can be found in manufacturing plants,
construction sites, automotive workshops, and even in
households.

4. COMMON USES OF AIR COMPRESSORS
Powering Pneumatic tools
- Compressed air is used to operate tools such as impact
wrenches, nail guns, spray guns, and air drills.
Inflating tires and sports equipment
- Air compressors can quickly inflate tires, sports balls, inflatable
toys, and other objects.
HVAC systems
- Air compressors play a crucial role in air conditioning and
refrigeration systems by compressing refrigerant gases.
Industrial processes
- Compressed air is used in various industrial processes, including
powering machinery, controlling valves and actuators, and
providing air for ventilation and breathing systems.

5. Classification
Air Compressors
Reciprocating Rotary
Single – acting
Double - Acting
No. of Sides of Piston
in operation
No. of Stages
for Compression
Centrifugal
Single – stage
Multi - stage

6. Classification
Reciprocating Air Compressors
- Reciprocating or piston compressors are one of the most
common types. They use a piston-cylinder mechanism to
compress air. The piston moves back and forth within a cylinder,
creating a vacuum on the intake stroke and compressing the air
on the compression stroke.
Rotary compressor
- also known as a rotary screw compressor, is a type of air
compressor that uses a rotary mechanism to compress air. It is
widely used in various industrial applications due to its efficiency,
reliability, and continuous operation

7. Reciprocating Compressor - Working

8. Reciprocating Compressor - Working
Principle of Operation
 Fig. shows single-acting piston actions in the
cylinder of a reciprocating compressor.
 The piston is driven by a crank shaft via a
connecting rod.
 At the top of the cylinder are a suction valve
and a discharge valve.
 A reciprocating compressor usually has two,
three, four, or six cylinders in it.

9. Reciprocating Compressor – Equation for Work
Volume
Pressure
P1
P2
V1
V2
3 2 2”
2’
4 1 (Polytropic)
(Adiabatic)
(Isothermal)
C
V
P n

C
V
P 

C
V
P 
Operations :
4 – 1 : Volume V1 of air aspirated into Compressor, at P1 and T1.
1 – 2 : Air compressed according to PVn = Const. from P1 to P2.
→ Temp increase from T1 to T2.
2 – 3 : Compressed air at P2 and V2 with temperature T2 is delivered.

10. Reciprocating Compressor – Equation for Work
During Compression, due to the excess temperature above
surrounding, the air will exchange the heat to the surrounding.
 Compression Index, n is always less than γ, the adiabatic index.
As Compressor is a work consuming device, every effort is desired to
reduce the work.
Work done = Area under P-V curve
 1 – 2” : Adiabatic Compression = Max. Work.
 1 – 2 : Polytropic Compression
 1 – 2’ : Isothermal Compression = Min. Work.

11. Reciprocating Compressor – Equation for Work
Thus, comparison between the Isothermal Work and the Actual
Work is important.
Isothermal Efficiency, ηiso =
Isothermal Work
Actual Work
Thus, more the Isothermal Efficiency, more the actual compression
approaches to the Isothermal Compression.
P1
P2
V1
V2
3 2 2”
2’
4 1(Polytropic)
(Adiabatic)
(Isothermal)
C
V
P n

C
V
P 

C
V
P 
Actual Work = Wact = Area 4-1-2-3-4
Wact = Area (4-1) – Area (1-2) – Area (2-3)
 
  


























1
1
1
2
2
1
1
2
2
1
1
1
1
2
2
2
2
1
1
2
2
1
1
2
2
1
1
n
V
P
V
P
V
P
V
P
n
V
P
V
P
V
P
V
P
V
P
n
V
P
V
P
V
P

12.  
 




































1
1
2
2
1
1
2
2
1
1
2
2
1
1
1
1
1
1
1
1
V
P
V
P
V
P
n
n
V
P
V
P
n
n
V
P
V
P
n
Wiso
Reciprocating Compressor – Equation for Work
P1
P2
V1
V2
3 2 2”
2’
4 1(Polytropic)
(Adiabatic)
(Isothermal)
C
V
P n

C
V
P 

C
V
P 
Now,
n
n
n
P
P
V
V
V
P
V
P
/
1
2
1
1
2
2
2
1
1






































n
iso
P
P
P
P
V
P
n
n
W
/
1
2
1
1
2
1
1 1
1

13. 





















































 n
n
iso
P
P
P
P
V
P
n
n
P
P
P
P
V
P
n
n
W
/
1
1
2
1
2
1
1
/
1
2
1
1
2
1
1
1
1
1
1
Reciprocating Compressor – Equation for Work
P1
P2
V1
V2
3 2 2”
2’
4 1(Polytropic)
(Adiabatic)
(Isothermal)
C
V
P n

C
V
P 

C
V
P 




























n
n
iso
P
P
V
P
n
n
W
1
1
2
1
1 1
1
The solution of this equation is always negative.
This shows that Work is done ON the Compressor.




























n
n
iso
P
P
mRT
n
n
W
1
1
2
1 1
1
Delivery Temperature,
n
n
P
P
T
T
1
1
2
1
2











14. Reciprocating Compressor – Equation for Work
P1
P2
V1
V4
6 2
5 1
C
V
P n

3
4
V3
Effective Swept Volume, V1-V4
Swept Volume, V1-V3=Vs
Total Volume, V1
Clearance Volume,
V3=Vc
Clearance Volume :
Volume that remains inside the cylinder
after the piston reaches the end of its
inward stroke.
Thus, Effective Stroke Volume = V1 – V4
Actual Work
Wact = Area 1-2-3-4
Wact = Area (5-1-2-6) – Area (5-4-3-6)

15. Reciprocating Compressor – Equation for Work
 




























n
act
P
P
P
P
V
V
P
n
n
W
/
1
2
1
1
2
4
1
1 1
1












































n
m
n
m
act
P
P
V
P
n
n
P
P
V
P
n
n
W
1
4
3
4
4
1
1
2
1
1 1
1
1
1
P1
P2
V1
V4
6 2
5 1
C
V
P n

3
4
V3
Effective Swept Volume, V1-V4
Swept Volume, V1-V3=Vs
Total Volume, V1
Clearance
Volume,
V3=Vc
But, P4 = P1 and P3 = P2

16. Reciprocating Compressor – Volumetric Efficiency
Volumetric Efficiency :
Ratio of free air delivered to the displacement of the compressor.
Ratio of Effective Swept Volume to Swept Volume.
Volumetric Efficiency =
Effective Swept Volume
Swept Volume
V1 – V4
V1 – V3
=
Vc
Vs
= = γ
Clearance Volume
Swept Volume
Clearance Ratio =
Presence of Clearance Volume
Volumetric Efficiency less than 1. ( 60 – 85 % )
P1
P2
V1
V4
6 2
5 1
C
V
P n

3
4
V3
Effective Swept Volume, V1-V4
Swept Volume, V1-V3=Vs
Total Volume, V1
Clearance Volume,
V3=Vc
( 4 – 10 % )

17. Reciprocating Compressor – Volumetric Efficiency

↑ Pr. Ratio ↑ Effect of Clearance Volume
….Clearance air expansion through greater volume before intake


Cylinder bore and stroke is fixed.
Effective Swept Volume (V1 – V4) ↓ with ↑ Pr. Ratio
↓ Volumetric Efficiency


   
     
   
    3
4
3
1
3
3
1
3
3
3
3
1
4
3
1
3
3
1
4
3
1
3
3
1
4
3
3
1
3
1
4
1
1
1
1
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
vol


























P1
P2
V1
V4
6 2
5 1
3
4
V3
Effective Swept Volume,
V1-V4
Swept Volume, V1-V3=Vs
Total Volume, V1
Clearance Volume,
V3=Vc

18. Reciprocating Compressor – Volumetric Efficiency
































































1
1
1
1
1
1
1
1
/
1
4
3
/
1
4
3
3
1
3
4
3
3
1
3
4
3
3
1
3
n
vol
n
vol
vol
vol
P
P
P
P
V
V
V
V
V
V
V
V
V
V
V
V
V





P1
P2
V1
V4
6 2
5 1
3
4
V3
Effective Swept Volume,
V1-V4
Swept Volume, V1-V3=Vs
Total Volume, V1
Clearance Volume,
V3=Vc

19. Reciprocating Compressor – F.A.D.
Free Air Delivery (F.A.D.) : If the volume of the air compressor is reduced to atmospheric
temperature and pressure, this volume of air is called FAD (m3/min).
Delivered mass of air = intake mass of air
   
2
3
2
2
1
4
1
1
T
V
V
P
T
V
V
P
T
V
P
t
t
t 



If clearance volume is neglected
Where
K
C
T
m
KN
P
t
t
288
15
/
325
.
101
0
2



2
2
2
1
1
1
T
V
P
T
V
P
T
V
P
t
t
t



20. Reciprocating Compressor – Single Stage
High Pressure required by Single – Stage :
1. Requires heavy working parts.
2. Has to accommodate high pressure ratios.
3. Increased balancing problems.
4. High Torque fluctuations.
5. Requires heavy Flywheel installations.

21. Reciprocating Compressor – Multistage
Series arrangement of cylinders, in which the compressed air
from earlier cylinder (i.e. discharge) becomes the intake air for the
next cylinder (i.e. inlet).
Intercooler :
Compressed air is cooled
between cylinders.
L.P. = Low Pressure
I.P. = Intermediate
Pressure
H.P. = High Pressure
L.P.
Cylinder
I.P.
Cylinder
H.P.
Cylinder
Intercooler
Intercooler
Air Intake
Air Delivery

22. Reciprocating Compressor – Multistage
Intake Pr.
P1 or Ps
Delivery Pr.
P3 or Pd
3
2
9 5
4
1
C
V
P n

8
7
6
Intermediate Pr.
P2 C
V
P 
Without Intercooling
Perfect Intercooling
L.P.
H.P.
Volume
Overall Pr. Range : P1 – P3
Single – stage cycle : 8-1-5-6
Without Intercooling :
L.P. : 8-1-4-7
H.P. : 7-4-5-6
With Intercooling :
L.P. : 8-1-4-7
H.P. : 7-2-3-6
Perfect Intercooling : After initial compression in L.P. cylinder, air is cooled in the
Intercooler to its original temperature, before entering H.P. cylinder
i.e. T2 = T1 OR
Points 1 and 2 are on SAME Isothermal line.

23. Reciprocating Compressor – Single Stage
Ideal Conditions for Multi – Stage Compressors
A. Single – Stage Compressor :
C
V
P 
3
2
9 5
4
1
C
V
P n

8
7
6
L.P.
H.P.
Single – stage cycle : 8-1-5-6





















1
1
5
1
1 1
1
n
n
P
P
V
P
n
n
W
Delivery Temperature,
n
n
P
P
T
T
1
1
5
1
5











24. Reciprocating Compressor – Multistage
C
V
P 
3
2
9 5
4
1
C
V
P n

8
7
6
L.P.
H.P.
B. Two – Stage Compressor (Without Intercooling) :
Without Intercooling :
L.P. : 8-1-4-7
H.P. : 7-4-5-6












































n
n
n
n
P
P
V
P
n
n
P
P
V
P
n
n
W
1
4
5
4
4
1
1
4
1
1
1
1
1
1
This is SAME as that of Work done in Single – Stage.
Delivery Temperature also remains SAME.
Without Intercooling 

25. Reciprocating Compressor – Multistage
C
V
P 
3
2
9 5
4
1
C
V
P n

8
7
6
L.P.
H.P.
C. Two – Stage Compressor (With Perfect Intercooling) :
With Intercooling :
L.P. : 8-1-4-7-8
H.P. : 7-2-3-6-7












































n
n
n
n
P
P
V
P
n
n
P
P
V
P
n
n
W
1
2
3
2
2
1
1
4
1
1
1
1
1
1
Delivery Temperature,
1
2
1
2
3
1
1
2
3
2
3 , T
T
as
P
P
T
P
P
T
T
n
n
n
n






















26. Reciprocating Compressor – Multistage
C
V
P 
3
2
9 5
4
1
C
V
P n

8
7
6
L.P.
H.P.
C. Two – Stage Compressor (With Perfect Intercooling) :
With Intercooling :
L.P. : 8-1-4-7-8
H.P. : 7-2-3-6-7
































n
n
n
n
P
P
P
P
V
P
n
n
W
1
2
3
1
1
2
1
1 2
1
Now, T2 = T1
P2V2 = P1V1
Also P4 = P2
Shaded Area 2-4-5-3-2 : Work Saving due to Intercooler…!!

27. Reciprocating Compressor – Multistage
Condition for Min. Work :
C
V
P 
3
2
9 5
4
1
C
V
P n

8
7
6
L.P.
H.P.
Intermediate Pr. P2 → P1 : Area 2-4-5-3-2 → 0
Intermediate Pr. P2 → P3 : Area 2-4-5-3-2 → 0
 There is an Optimum P2 for which Area 2-4-5-3-2
is maximum,
i.e. Work is minimum…!!
































n
n
n
n
P
P
P
P
V
P
n
n
W
1
2
3
1
1
2
1
1 2
1
0
2
1
2
3
1
1
2
2































dP
P
P
P
P
d
dP
dW
n
n
n
n
For min. Work,

28. Reciprocating Compressor – Multistage
Condition for Min. Work :
 
      0
1
1
1 1
1
2
1
3
1
1
2
1
1












 








 







 






 






 

n
n
n
n
n
n
n
n
P
n
n
P
P
n
n
P
0
2
1
2
3
1
1
2
2































dP
P
P
P
P
d
dP
dW
n
n
n
n
 
 
  




 





 


 n
n
n
n
n
P
P
P
P 1
3
1
1
2
2
/
1
2
   
3
1
2
2 P
P
P 
2
3
1
2
3
1
2
P
P
P
P
OR
P
P
P 

C
V
P 
3
2
9 5
4
1
C
V
P n

8
7
6
L.P.
H.P.

29. Reciprocating Compressor – Multistage
P2 obtained with this condition (Pr. Ratio per stage is equal) is the Ideal
Intermediate Pr. Which, with Perfect Intercooling, gives Minimum
Work, Wmin.
 






















n
n
P
P
P
V
P
n
n
W
1
1
2
/
1
3
1
1
1 1
1
2






















n
n
P
P
V
P
n
n
W
1
1
2
1
1 1
1
2























n
n
P
P
V
P
n
n
W
2
1
1
3
1
1 1
1
2
Equal Work per cylinder

30. Reciprocating Compressor – Efficiency
Isothermal work done / cycle = Area of P – V Diagram
= P1V1 loge(P2/P1)
Isothermal Power = P1V1 loge(P2/P1) N
60 X 1000
kW
Indicated Power : Power obtained from the actual indicator card taken during a
test on the compressor.
Compressor Efficiency = Isothermal Power
Indicated Power
Isothermal Efficiency = Isothermal Power
Shaft Power
NOTE : Shaft Power = Brake Power required to drive the Compressor.

31. Reciprocating Compressor – Efficiency
Adiabatic Efficiency : Ratio of Power required to drive the Compressor;
compared with the area of the hypothetical Indicator Diagram; assuming
Adiabatic Compression.
Compressor
the
drive
to
required
Power
Brake
P
P
V
P
adiabatic



























1
1
2
1
1 1
1
Mechanical Efficiency : Ratio of mechanical output to mechanical input.
Mechanical Efficiency, ηmech = Indicated Power
Shaft Power

32. Problem 1.0
A single acting air compressor operates at 150 rpm with an initial
condition of air at 97.9 Kpa and 27℃ and discharges the air at 379 Kpa
to a cylindrical tank. The bore and stroke are 355 mm and 381 mm,
respectively, with 5% clearance. If the surrounding air is at 100 Kpa
and 20℃ while the compression and expansion process are PV1.3 = C,
determine free air capacity, m3/sec.
Given:
150 rpm - N
97.9 Kpa – P1
379 Kpa – P2
100 kPa – P3
27℃ - T1
20 ℃ - T2
355 mm - 0.355 m – D1
381 mm – 0.381 m - D2
5% - 0.05 - c
Required:
• free air capacity, m3/sec.

33. Problem 1.0
Solution:
VD =
𝜋
4
D2 L N
VD =
𝜋
4
(0.355)2 (0.381) (150/60)
VD = 0.094278 m3/sec
ηV = 1 + c – c (P2/P1)1/n
ηV = 1 + 0.05 – 0.05 (379/97.9)1/1.3
ηV = 0.908
V1 = 0.908 (0.094278)
V1 = 0.085604 m3 / sec
Solving for free air capacity:
PFVF/TF = P1V1/T1
100(VF) / (20+273) = 97.9(0.085604) / (27+273)
VF = 0.081851 m3 / sec

34. Problem 2.0
A single acting reciprocating air compressor has a clearance volume of
10%. Air is received at 90 Kpa and 29.3 ℃ and is discharged at 600
Kpa. The compression and expansion are polytropic with n = 1.28. The
pressure drop is 5 Kpa at suction port and 10 Kpa at the discharged
port. The compressor piston displacement is 500 cm3 when operating at
900 rpm. Determine the mass of compressed air in kg/hr.
Given:
10% - 0.1 - c
90 Kpa – P1
600 Kpa – P2
5 Kpa – P3
10 Kpa - P4
29.3 ℃ - T1
20 ℃ - T2
500 cm3 - D1
900 rpm - N
1.28 - n
Required:
• mass of compressed air
in kg/hr.

35. Problem 2.0
Solution:
VD = (𝜋/4 D2) LN
VD = (500)(900)
VD = 450,000 cm3 / min
VD = 0.45 m3/min
P1 = 90 – 5
P1 = 85 Kpa
P2 = 600 + 10
P2 = 610 Kpa
ηV = 1 + c – c (P2/P1)1/n
η V = 1 + 0.10 – 0.10 (610/85)1/1.28
η V = 0.633684
V1 = 0.45 (0.633684)
V1 = 0.285 m3 / min
m = PV/RT
m = 85(0.285)/(0287)(29.3+273)
m= 0.2792 kg/min
m = 16.76 kg/hr

36. Problem 3.0
A two stage compressor air at 100 Kpa and 22 ℃ discharges to 690 Kpa.
If intercooler intake is 105℃, determine the value of n.
Given:
100 Kpa – P1
690 Kpa – P1
22 ℃ - T1
105 ℃ - T2
Required:
• value of n
Solution:
PX = 𝑃1𝑃2
PX = 100 690
PX = 262.68 Kpa
TX / T1 = (PX /P1) n-1/n
(105+273) / (22 + 273) = (262.68 / 100) n-1/n
1.281 = (2.6268) n-1/n
n- 1 / n = ln(1.281) / ln(2.6268)
n-1 = 0.2564 n
n = 1.345